Given that $1 {c m}^{3}$ of a metal weighs $5 g m s$ , the weight of a cylindrical metal container with base radius $20.5 c m$ and height $60 c m$ is:

976.5 k g

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487.5 k g

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396.24 k g

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102.45 k g

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Solution

The correct option is C $396.24 k g$
The radius of base of container $= 20.5 c m$
Height of the container $= 60 c m$
Volume of cylindrical container $= π r^{2} h = \frac{22}{7} \times 20.5 \times 20.5 \times 60$ $= 79247.14 c m^{3}$
Weight of $1 c m^{3}$ metal $= 5 g m s$
Weight of $79247.14 c m^{3}$ metal $= 5 \times 79247.14 = 396235.7 g m = 396.24 k g$

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Given that 1cm3 of a metal weighs 5gms, the weight of a cylindrical metal container with base radius 20.5cm and height 60cm is:

The correct option is C 396.24kgThe radius of base of container =20.5cmHeight of the container =60cm Volume of cylindrical container =πr2h=227×20.5×20.5×60=79247.14cm3Weight of 1cm3 metal =5gmsWeight of 79247.14cm3metal =5×79247.14=396235.7 gm =396.24 kg

Given that $1 {c m}^{3}$ of a metal weighs $5 g m s$ , the weight of a cylindrical metal container with base radius $20.5 c m$ and height $60 c m$ is: